Recent content by Rafa54G2

Wow, tried hundreds of things and nothing... Next time I will use logic instead of maths. Thank you! I don't know if you're a teacher, but you would be a good one, didn't solve the problem but helping me to do it :D I guess that b) is the same but changing the inflexion point and being carefull...

The possible point is maybe (-1,-5) considering G as (0,0)?

About the center of the base? When I said that crane would tip when G was outside the base, I meant this:

I've just read that Pisa's tower won't fall down because its mass center is in the area of the base. So, if G is outside the base, crane will start to tip. It depends on the angle formed by G and the cener of the crane base. Then, the distances measuared in the torque will change, and the mass X...

Well, without any up force, the crane will fall easily. What about consider an up force in the base of the crane? Now the system is in equilibrium because the module of the new force is the sum of the others, isn't it?

I'm considering 3 forces, Load force, G force and counterweight force. Should I consider more? The distances start in the crane axis. Torque= 10.5(m)*4(T)*9.8(m/s²) + 0.5(m)*10(T)*9.8(m/s²) - 2.25(m)*x(T)*9.8(m/s²) =0

Homework Statement Hello, this problem is cracking my mind for days, and I haven't solved it yet :( Crane dimensions: See image Mass of the crane: 10Tons Mass center: G Maximum Load: 4Tons Counterweight: P Weight of the counterweight: ? "A" says: when you have a maximum load (4Tons) at the...